Conic Sections — JEE MCQs with answers

Answer: A — $a^2=100$ so $a=10$, and the length of the major axis is $2a=20$ -- Option A. One option gives $10$, which is the semi-major axis $a$ (and also $2b$, the minor-axis length) -- neither is the major axis. One option gives $2c=10\sqrt{3}$, the distance between the two foci. One option gives $100$, the denominator $a^2$ itself, with neither the square root nor the doubling carried out.

Answer: A — Option A is correct: “The intersection of the plane with the cone can take place either at the vertex of the cone or at any other part of the nappe.” That is exactly two broad locations — at the vertex (giving the degenerate cases of §10.2.2), or elsewhere on a nappe (giving the non-degenerate cases of §10.2.1). Restricting it to the vertex alone drops every one of the §10.2.1 curves. The text does add “either below or above the vertex”, but it never treats below and above as separate cases — both give the same four sections — so there are not three. And §10.2 counts locations of the cut, not conic sections, so four is counting the wrong thing.

Answer: A — Option A is correct. Here $a^2=16$ so $a=4$, and $b^2=1$. The length of the latus rectum is $\dfrac{2b^2}{a}=\frac{1}{2}$. Option B is wrong: using $b^2/a$ (half the correct value) instead of $2b^2/a$. Option C is wrong: using $b^2/(4a)$ (an eighth of the correct value) instead of $2b^2/a$. Option D is wrong: using $b^2/(2a)$ (a quarter of the correct value) instead of $2b^2/a$.

Answer: B — A circle $(x - h)^2 + (y - k)^2 = r^2$ passes through the origin exactly when $(0 - h)^2 + (0 - k)^2 = r^2$, i.e. $h^2 + k^2 = r^2$. As a check, the circle with centre $(3, 4)$ and radius 5 passes through the origin, and $9 + 16 = 25$. Option A is incorrect: it expands $(0 - h)^2$ and $(0 - k)^2$ as $-h^2$ and $-k^2$, a sign slip in squaring, giving $-h^2 - k^2 = r^2$. For the centre $(3, 4)$ and radius 5 this option gives $-25 = 25$, which is false. Option B is correct: $h^2 + k^2 = r^2$. Option C is incorrect: it makes two slips at once: the sign slip in squaring $(0 - h)^2$ and $(0 - k)^2$, and writing $r$ instead of $r^2$, giving $-h^2 - k^2 = r$. For the centre $(3, 4)$ and radius 5 this option gives $-25 = 5$, which is false. Option D is incorrect: it drops the square on the radius and writes $r$ instead of $r^2$, giving $h^2 + k^2 = r$. For the centre $(3, 4)$ and radius 5 this option gives $25 = 5$, which is false.

Answer: B — Comparing $x^2 = -12y$ with the standard form $x^2 = -4ay$ (pp.183-184) gives $a = 3$, so the focus is $(0, -3)$ and the directrix is the line $y = 3$ on the other side of the vertex. Every point of the parabola is equidistant from the focus and the directrix (Definition 2); the point $(-6, -3)$ lies on the parabola and is at squared distance $36$ from the focus. Option A is incorrect: it uses $2a$ in place of $a$, giving $y = 6$; the point $(-6, -3)$ is at squared distance $81$ from this line, not $36$, so it cannot be the directrix. Option B is correct: $y = 3$, which is also at squared distance $36$ from that point. Option C is incorrect: it makes two slips at once: flips the sign of the constant, putting the line on the same side of the vertex as the focus, and uses $2a$ in place of $a$, giving $y = -6$; the point $(-6, -3)$ is at squared distance $9$ from this line, not $36$, so it cannot be the directrix. Option D is incorrect: it flips the sign of the constant, putting the line on the same side of the vertex as the focus, giving $y = -3$; the point $(-6, -3)$ is at squared distance $0$ from this line, not $36$, so it cannot be the directrix.

Answer: C — Option A is wrong because §10.1 never corrects the 'preceding Chapter 10' error; this closing sentence is about the plane-cone intersection and is unrelated to that erratum. Option B is wrong because eccentricity is introduced only in §10.5 and §10.6, and nothing about it is established before this closing sentence of §10.1. Option C is correct because the sentence's own wording — 'in the subsequent sections we will see...' — signals that it looks ahead, preparing the reader for §10.2 onward, rather than defining or summarising anything already given. Option D is wrong because Definition 1 appears later, in §10.3, and is not restated here.

Answer: C — $b^2=a^2-c^2=169-25=144$ -- Option C. One option adds instead of subtracting, $a^2+c^2=194$ (the hyperbola relation). One option sets $b^2=c^2=25$, mistaking the focal distance for the minor semi-axis. One option computes $c^2-a^2=-144$, which is negative and so cannot be a squared length.

Answer: B — Option B is correct: “the line l is the axis of the cone.” “The generator” is the rotating line m, not the fixed line l. “The vertex” is the point V, not a line. “A nappe” is one of the two parts of the cone's surface, not a line.

Answer: A — Option A is correct. The transverse axis has length $2a=16$, so $a=8$, $a^2=64$. Since $c=17$, $b^2=c^2-a^2=225$. The equation is $\dfrac{x^2}{64} - \dfrac{y^2}{225} = 1$. Option B is wrong: using $a=16$ (the full transverse-axis length) instead of $a=8$; this gives $a^2=256$ and $b^2=289-256=33$, a positive but incorrect pair, because the transverse axis has length $2a$, not $a$. Option C is wrong: placing the transverse axis on the wrong coordinate axis. Option D is wrong: treating the given length as the conjugate axis ($2b$) instead of the transverse axis ($2a$).

Answer: C — Substituting the centre (h, 0) into $|CP| = r$ for $P = (8, 15)$ gives the quadratic $h^2 - 2(8)h + (0) = 0$, whose two roots are the two possible x-coordinates of the centre; by Vieta their sum is minus the linear coefficient, i.e. $2 \times 8 = 16$ -- the key. $-16$ comes from expanding $(x_1 - h)^2$ as if it were $(x_1 + h)^2$, giving a quadratic whose roots sum to $-16 \neq 16$. $8$ comes from expanding $(x_1 - h)^2$ so the linear coefficient is missing its factor of 2 (uses $-x_1h$ instead of $-2x_1h$), giving a quadratic whose roots sum to $8 \neq 16$. $0$ comes from expanding $(x_1 - h)^2$ as $x_1^2 - h^2$ (dropping the cross term $-2x_1h$), so the linear term vanishes, giving a quadratic whose roots sum to $0 \neq 16$.